limit, challenge problem

[lookagain]

I read this problem in a book or a magazine. It consists of infinitely nested square roots.


The following limit does exist. \(\displaystyle \ \ \) Find the limit.



\(\displaystyle \displaystyle\lim_{x\to 0^{+}} \ \sqrt{x \ + \ \sqrt{x \ + \ \sqrt{x \ + \ \sqrt{x \ + \ ... \ } \ } \ } \ }\)

 

[daon2]

In order to treat a limiting value as a number, we must show convergence. I claim the assignment \(\displaystyle f(x)=\sqrt{x+\sqrt{x+\cdots}}\) converges for all \(\displaystyle x>0\), but I will show for \(\displaystyle x<1\), using the monotone convergence theorem.

To have this we need to show that the sequence defined by \(\displaystyle a_0(x)=\sqrt{x}, a_n(x) = \sqrt{x+a_{n-1}}\) converges. For short, call them \(\displaystyle a_0, a_n\). Well, first \(\displaystyle a_n\) is non-decreasing. Certainly \(\displaystyle a_0=\sqrt{x} < \sqrt{x+\sqrt{x}} = a_1\). Now, \(\displaystyle a_n^2 = x+a_{n-1} \le x+a_{n} = a_{n+1}^2\) and hence \(\displaystyle a_n < a_{n+1}\).

Finally, we need to show \(\displaystyle a_n\) is bounded. I claim \(\displaystyle a_n\le 3\), which is clearly true for \(\displaystyle a_0\). Now, suppose \(\displaystyle a_{n-1}\le 3\). \(\displaystyle a_n = \sqrt{x+a_{n-1}} \le \sqrt{1+3} = 2 < 3\). Therefore \(\displaystyle a_n\) converges, and by construction has limit equal to \(\displaystyle f(x)\), i.e., \(\displaystyle f(x) = \lim_{n\to \infty} a_n \).

The question now comes down to the answer of \(\displaystyle \lim_{x\to{0^+}} f(x)\). Using the relation \(\displaystyle f(x) = \sqrt{x+f(x)}\), let's examine the equation \(\displaystyle f(x)^2 - f(x) - x=0\). As a "quadratic" we have the roots must satisfy:

\(\displaystyle f(x) = \dfrac{1}{2}\left(1\pm \sqrt{4x+1}\right)\).

Let us assume first that the "plus" solution is the correct one. Taking the limit as \(\displaystyle x\to 0^{+}\), we see that

\(\displaystyle \lim_{x\to 0^{+}}f(x) = \lim_{x\to 0^+} \dfrac{1}{2}\left(1+\sqrt{4x+1}\right) = \dfrac{1}{2}(1+1)=1\).

Now we shall see that the limit cannot be zero, by showing the "minus" solution is incorrect, so we pretend it is correct. Assume that \(\displaystyle c\) is a number verrrrrrrrry close to, but bigger than, zero. We have that \(\displaystyle 4c+1 > 1\), and hence \(\displaystyle f(c) = \dfrac{1-\sqrt{4c+1}}{2} < 0\). But that's impossible, since \(\displaystyle f(c) \ge a_n(c) \ge \sqrt{c} > 0\).

Therefore

\(\displaystyle \lim_{x\to 0^{+}}f(x)=\lim_{x\to 0^{+}}\sqrt{x+\sqrt{x+\cdots}} = 1\).